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What type of equation is this one?

  1. Apr 27, 2014 #1
    Hi,

    I am struggling with a problem, where in the middle of my calculations I need to determine a function ##f(\alpha, \beta; x)##, namely the function of ##x## parametrized by ##\alpha## and ##\beta##, from the following equation

    $$f\left(\alpha, \beta;\frac{1}{x}\right) = x^4f(\gamma, \delta; x).$$

    So the ##f(1/x)## on the LHS and ##f(x)## on the RHS may differ in parametrization. Somehow I found that this equation admits the following solution

    $$f(\alpha, \beta; x) = \left(\alpha + \beta x^a\right)^b $$

    where ##a\cdot b = -4##. Although, this is probably not the most general solution.

    So my question is what type of equation this one is? And how one should approach such equation?
     
    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2

    disregardthat

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    Science Advisor

    Some clarifying questions:

    Is x a non-zero real number?

    Are [itex]\alpha, \beta[/itex] real numbers?

    Are [itex]\gamma[/itex] and [itex]\delta[/itex] real number constants (independent of [itex]\alpha, \beta, x[/itex])?

    In your example, you're assuming [itex]\gamma = \alpha = \delta = \beta[/itex]?
     
  4. Apr 27, 2014 #3
    ##f(x)## is well defined and restricted to either ##x>0## or ##x<0##.
    ##\alpha##, ##\beta##, ##\gamma##, ##\delta## in general may be complex and independent of ##x## (they parametrize function ##f(x)##).
    However ##\gamma##, ##\delta## may be functions of ##\alpha##, ##\beta##.

    In the example I provided previously we have

    $$f\left(\alpha, \beta; x\right) = \left(\alpha + \beta x^{a} \right)^b \Rightarrow f\left(\alpha, \beta; \frac{1}{x}\right) = \left(\alpha + \beta x^{-a} \right)^b = x^{-ab}\left(\alpha x^a + \beta\right)^b = x^{4}f(\beta, \alpha; x)$$
    Where the assumption ##ab = -4## was used. So in this case ##\alpha = \delta## and ##\beta = \gamma## and are not neccesarily equal.

    Sorry for not making this clear enough earlier.
     
    Last edited: Apr 27, 2014
  5. May 4, 2014 #4
    Anyone?
     
  6. May 4, 2014 #5
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